05 Parametric Surfaces.pdf

(1)

Parametric Surfaces and Surfaces of Revolution

Lucky Galvez

Institute of Mathematics University of the Philippines

Diliman

(2)

Recall

A curve in

R

3

is given by a vector function

R(

t

) =

f

(

t

)ˆ

ı

+

g

(

t

)ˆ



+

h

(

t

)ˆ

k

or a set of parametric equations

x

=

f

(

t

)

,

y

=

g

(

t

)

,

z

=

h

(

t

)

.

(3)

Parametric Surfaces

A surface in

R

3

can be described by a vector function of two

parameters

R(

u, v

).

Suppose that

R(

u, v

) =

x

(

u, v

)ˆ

ı

+

y

(

u, v

)ˆ



+

z

(

u, v

)ˆ

k

is a vector function defined on a parameter domain

D

(in the

uv

-plane). Any particular choice of (

u, v

)

∈

D

gives a point

(

x, y, z

) such that

x

=

x

(

u, v

)

y

=

y

(

u, v

)

z

=

z

(

u, v

)

.

The set of all such points as (

u, v

) varies over

D

is called a

parametric surface.

The parametric surface is

traced out by the tip of the

moving vector

R(

u, v

) as (

u, v

)

varies over

D

.

(4)

Parametric Surfaces

A surface in

R

3

parameters

R(

u, v

). Suppose that

R(

u, v

) =

x

(

u, v

)ˆ

ı

+

y

(

u, v

)ˆ



+

z

(

u, v

)ˆ

k

D

(in the

uv

-plane).

Any particular choice of (

u, v

)

∈

D

gives a point

(

x, y, z

) such that

x

=

x

(

u, v

)

y

=

y

(

u, v

)

z

=

z

(

u, v

)

.

u, v

) varies over

D

is called a

parametric surface.

The parametric surface is

moving vector

R(

u, v

) as (

u, v

)

varies over

D

.

(5)

Parametric Surfaces

A surface in

R

3

parameters

R(

u, v

). Suppose that

R(

u, v

) =

x

(

u, v

)ˆ

ı

+

y

(

u, v

)ˆ



+

z

(

u, v

)ˆ

k

D

(in the

uv

u, v

)

∈

D

gives a point

(

x, y, z

) such that

x

=

x

(

u, v

)

y

=

y

(

u, v

)

z

=

z

(

u, v

)

.

u, v

) varies over

D

is called a

parametric surface.

moving vector

R(

u, v

) as (

u, v

)

varies over

D

.

(6)

Parametric Surfaces

A surface in

R

3

parameters

R(

u, v

). Suppose that

R(

u, v

) =

x

(

u, v

)ˆ

ı

+

y

(

u, v

)ˆ



+

z

(

u, v

)ˆ

k

is a vector function defined on a parameter domain

D

(in the

uv

u, v

)

∈

D

gives a point

(

x, y, z

) such that

x

=

x

(

u, v

)

y

=

y

(

u, v

)

z

=

z

(

u, v

)

.

u, v

) varies over

D

is called a

parametric surface.

moving vector

R(

u, v

) as (

u, v

)

varies over

D

.

(7)

Parametric Surfaces

A surface in

R

3

parameters

R(

u, v

). Suppose that

R(

u, v

) =

x

(

u, v

)ˆ

ı

+

y

(

u, v

)ˆ



+

z

(

u, v

)ˆ

k

D

(in the

uv

u, v

)

∈

D

gives a point

(

x, y, z

) such that

x

=

x

(

u, v

)

y

=

y

(

u, v

)

z

=

z

(

u, v

)

.

u, v

) varies over

D

is called a

parametric surface.

The parametric surface is

moving vector

R(

u, v

) as (

u, v

)

varies over

D

.

(8)

Parametric Surfaces

Example

Identify the surface given by

R(

u, v

) =

h

u,

cos

v,

sin

v

i

.

Solution.

The corresponding set of parametric equations are

x

=

u,

y

= cos

v,

z

= sin

v.

Note that for any point (

x, y, z

) on the surface,

y

2

+

z

2

= cos

2

v

+ sin

2

v

= 1

This means that for constant

x

, the cross sections parallel to

the

x

-axis are circles of radius 1.

Hence, the surface is a right

circular cylinder

(9)

Parametric Surfaces

Example

R(

u, v

) =

h

u,

cos

v,

sin

v

i

.

Solution.

x

=

u,

y

= cos

v,

z

= sin

v.

x, y, z

) on the surface,

y

2

+

z

2

= cos

2

v

+ sin

2

v

= 1

x

the

x

Hence, the surface is a right

circular cylinder

(10)

Parametric Surfaces

Example

R(

u, v

) =

h

u,

cos

v,

sin

v

i

.

Solution.

x

=

u,

y

= cos

v,

z

= sin

v.

x, y, z

) on the surface,

y

2

+

z

2

= cos

2

v

+ sin

2

v

= 1

x

, the cross sections parallel to

the

x

circular cylinder

(11)

Parametric Surfaces

Example

R(

u, v

) =

h

u,

cos

v,

sin

v

i

.

Solution.

x

=

u,

y

= cos

v,

z

= sin

v.

x, y, z

) on the surface,

y

2

+

z

2

= cos

2

v

+ sin

2

v

= 1

x

the

x

circular cylinder

(12)

Parametric Surfaces

Example

R(

u, v

) =

h

u,

cos

v,

sin

v

i

.

Solution.

x

=

u,

y

= cos

v,

z

= sin

v.

x, y, z

) on the surface,

y

2

+

z

2

= cos

2

v

+ sin

2

v

= 1

x

the

x

Hence, the surface is a right

circular cylinder

(13)

Grid Curves

Consider a surface

S

given by a vector function

R(

u, v

).

If we holdu=u0, constant, then

R(u0, v) becomes a vector function of a single parameterv which traces a curveC1onS.

If we holdv=v0, constant, then

R(u, v0) becomes a vector function of a single parameteru which traces a curveC2onS.

We call these curves the

grid curves

of

S

.

(14)

Grid Curves

Consider a surface

S

R(

u, v

).

R(u0, v) becomes a vector function of a single parameterv

which traces a curveC1onS.

We call these curves the

grid curves

of

S

.

(15)

Grid Curves

Consider a surface

S

R(

u, v

).

We call these curves the

grid curves

of

S

.

(16)

Grid Curves

Consider a surface

S

R(

u, v

).

R(u, v0) becomes a vector function of a single parameteru

which traces a curveC2onS.

We call these curves the

grid curves

of

S

.

(17)

Grid Curves

Consider a surface

S

R(

u, v

).

We call these curves the

grid curves

of

S

.

(18)

Grid Curves

Consider a surface

S

R(

u, v

).

We call these curves the

grid curves

of

S

.

(19)

Grid Curves

The grid curves of

R(

u, v

) =

h

u,

cos

v

sin

v

i

:

(20)

Grid Curves

The grid curves of

R(

u, v

) =

h

u,

cos

v

sin

v

i

:

(21)

Grid Curves

The grid curves of

R(

u, v

) =

h

u,

cos

v

sin

v

i

:

(22)

Parametric Surfaces

Example

Describe the parametric surface given by the parametric

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

curves with constant

u

and grid curve with

v

= 0.

Solution.

To eliminate the parametersuandv note that x2+y2=u2cos2v+u2sin2v=u2,

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

z = 4−x2−y2, a paraboloid.

Ifu=u0, a constant, thenz= 4−u20 is constant and x2₊_y2₌_u2

0. Thus, we have circles parallel to the xy-plane.

Ifv= 0, the parametric equation becomes x=u, y= 0, z= 4−u2 which gives the parabolaz= 4−x2 _{on the}_xz-plane.

(23)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

To eliminate the parametersuandv note that x2+y2=u2cos2v+u2sin2v

=u2,

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(24)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(25)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_z.

Hence,

x2+y2 = u2= 4−z

(26)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2

= 4−z

(27)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(28)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

z = 4−x2−y2,

a paraboloid.

(29)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(30)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifu=u0, a constant,

thenz= 4−u2

0 is constant and x2₊_y2₌_u2

(31)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifu=u0, a constant, thenz= 4−u20 is constant and

x2₊_y2₌_u2

(32)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

0.

Thus, we have circles parallel to the xy-plane.

(33)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(34)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifv= 0, the parametric equation becomes

x=u, y= 0, z= 4−u2 which gives the parabolaz= 4−x2 _{on the}_xz-plane.

(35)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifv= 0, the parametric equation becomes x=u,

y= 0, z= 4−u2 which gives the parabolaz= 4−x2 _{on the}_xz-plane.

(36)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifv= 0, the parametric equation becomes x=u, y= 0,

z= 4−u2 which gives the parabolaz= 4−x2 _{on the}_xz-plane.

(37)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

Ifv= 0, the parametric equation becomes x=u, y= 0, z= 4−u2

which gives the parabolaz= 4−x2 _{on the}_xz-plane.

(38)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(39)

Parametric Surfaces

Example

equations

x

=

u

cos

v, y

=

u

sin

v, z

= 4

−

u

2

. Identify grid

u

and grid curve with

v

= 0.

Solution.

butz= 4−u2_⇒_u2_{= 4}₋_{z. Hence,}

x2+y2 = u2= 4−z

(40)

Surfaces of Revolution

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

be the surface obtained when

the curve

y

=

f

(

x

) is revolved about the

x

-axis.

Let

θ

be the

angle of rotation as shown below:

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

which is a parametric surface in the parameters

x

and

θ

.

(41)

Surfaces of Revolution

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

be the surface obtained when

the curve

y

=

f

(

x

-axis. Let

θ

be the

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

x

and

θ

.

(42)

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

the curve

y

=

f

(

x

-axis. Let

θ

be the

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

x

and

θ

.

(43)

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

the curve

y

=

f

(

x

) is revolved about the

x

-axis. Let

θ

be the

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

x

and

θ

.

(44)

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

the curve

y

=

f

(

x

-axis. Let

θ

be the

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

x

and

θ

.

(45)

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

the curve

y

=

f

(

x

-axis. Let

θ

be the

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

x

and

θ

.

(46)

Let

f

(

x

)

≥

0 for

a

≤

x

≤

b

and

S

the curve

y

=

f

(

x

-axis. Let

θ

be the

angle of rotation as shown below:

If (

x, y, z

) is a point on

S

, then

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ

(47)

If

S

is a surface obtained by revolving

y

=

f

(

x

) or

z

=

f

(

x

)

about the

x

-axis, then

S

has parametric equations

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ.

If

S

x

=

f

(

y

) or

z

=

f

(

y

)

about the

y

-axis, then

S

x

=

f(y) cos

θ,

y

=

y,

z

=

f(y) sin

θ.

If

S

x

=

f

(

z

) or

y

=

f

(

z

)

about the

z

-axis, then

S

x

=

f

(z) cos

θ,

y

=

f

(z) sin

θ,

z

=

z.

(48)

If

S

y

=

f

(

x

) or

z

=

f

(

x

)

about the

x

-axis, then

S

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ.

If

S

x

=

f

(

y

) or

z

=

f

(

y

)

about the

y

-axis, then

S

x

=

f(y) cos

θ,

y

=

y,

z

=

f(y) sin

θ.

If

S

x

=

f

(

z

) or

y

=

f

(

z

)

about the

z

-axis, then

S

x

=

f

(z) cos

θ,

y

=

f

(z) sin

θ,

z

=

z.

(49)

If

S

y

=

f

(

x

) or

z

=

f

(

x

)

about the

x

-axis, then

S

x

=

x,

y

=

f

(x) cos

θ,

z

=

f

(x) sin

θ.

If

S

x

=

f

(

y

) or

z

=

f

(

y

)

about the

y

-axis, then

S

x

=

f(y) cos

θ,

y

=

y,

z

=

f(y) sin

θ.

If

S

x

=

f

(

z

) or

y

=

f

(

z

)

about the

z

-axis, then

S

x

=

f

(z) cos

θ,

y

=

f

(z) sin

θ,

z

=

z.

(50)

Example

Give a set of parametric equations for the surface obtained by

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

1 The generating curve is y=f(x) =ex_.

Hence, the surface of revolution is given by

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

2 Note: y=ex⇒x= lny, so the generating curve is x=f(y) = lny. Hence, the surface of revolution is given by     

x= lnycosθ y= y

z= lnysinθ

(51)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

x= lnycosθ y= y

z= lnysinθ

(52)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

x= lnycosθ y= y

z= lnysinθ

(53)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

x= lnycosθ y= y

z= lnysinθ

(54)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

2 Note: y=ex⇒x= lny,

so the generating curve is x=f(y) = lny. Hence, the surface of revolution is given by     

x= lnycosθ y= y

z= lnysinθ

(55)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

2 Note: y=ex⇒x= lny, so the generating curve is x=f(y) = lny.

Hence, the surface of revolution is given by     

x= lnycosθ y= y

z= lnysinθ

(56)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

x= lnycosθ y= y

z= lnysinθ

(57)

Example

revolving

y

=

e

x

about the

1

x

-axis.

2

y

-axis.

Solution.

  

 

x= x y= ex_cos_θ

z= ex_sin_θ

x= lnycosθ y= y

z= lnysinθ

(58)

Tangent Plane to Parametric Surfaces

Consider a surface

S

given by

R(

u, v

) =

h

x

(

u, v

)

, y

(

u, v

)

, z

(

u, v

)

i

and a point

P

0

in

S

with position vector

R(

u

0

, v

0

).

If

C

1

is the grid curve obtained by setting

u

=

u

0

, then the

tangent vector to

C

1

at

P

0

is

R

v

(

u

0

, v

0

) =

h

x

v

(

u

0

, v

0

)

, y

v

(

u

0

, v

0

)

, z

v

(

u

0

, v

0

)

i

.

Similarly, if

C

2

v

=

v

0

,

then the tangent vector to

C

2

at

P

0

is

R

u

(

u

0

, v

0

) =

h

x

u

(

u

0

, v

0

)

, y

u

(

u

0

, v

0

)

, z

u

(

u

0

, v

0

)

i

.

(59)

Consider a surface

S

given by

R(

u, v

) =

h

x

(

u, v

)

, y

(

u, v

)

, z

(

u, v

)

i

and a point

P

0

in

S

R(

u

0

, v

0

).

If

C

1

u

=

u

0

, then the

tangent vector to

C

1

at

P

0

is

R

v

(

u

0

, v

0

) =

h

x

v

(

u

0

, v

0

)

, y

v

(

u

0

, v

0

)

, z

v

(

u

0

, v

0

)

i

.

Similarly, if

C

2

v

=

v

0

,

C

2

at

P

0

is

R

u

(

u

0

, v

0

) =

h

x

u

(

u

0

, v

0

)

, y

u

(

u

0

, v

0

)

, z

u

(

u

0

, v

0

)

i

.

(60)

Consider a surface

S

given by

R(

u, v

) =

h

x

(

u, v

)

, y

(

u, v

)

, z

(

u, v

)

i

and a point

P

0

in

S

with position vector

R(

u

0

, v

0

).

If

C

1

u

=

u

0

, then the

tangent vector to

C

1

at

P

0

is

R

v

(

u

0

, v

0

) =

h

x

v

(

u

0

, v

0

)

, y

v

(

u

0

, v

0

)

, z

v

(

u

0

, v

0

)

i

.

Similarly, if

C

2

v

=

v

0

,

C

2

at

P

0

is

R

u

(

u

0

, v

0

) =

h

x

u

(

u

0

, v

0

)

, y

u

(

u

0

, v

0

)

, z

u

(

u

0

, v

0

)

i

.

(61)

Consider a surface

S

given by

R(

u, v

) =

h

x

(

u, v

)

, y

(

u, v

)

, z

(

u, v

)

i

and a point

P

0

in

S

R(

u

0

, v

0

).

If

C

1

is the grid curve obtained by setting

u

=

u

0

, then the

tangent vector to

C

1

at

P

0

is

R

v

(

u

0

, v

0

) =

h

x

v

(

u

0

, v

0

)

, y

v

(

u

0

, v

0

)

, z

v

(

u

0

, v

0

)

i

.

Similarly, if

C

2

v

=

v

0

,

C

2

at

P

0

is

R

u

(

u

0

, v

0

) =

h

x

u

(

u

0

, v

0

)

, y

u

(

u

0

, v

0

)

, z

u

(

u

0

, v

0

)

i

.

(62)

If

R

u

×

R

v

6

= 0, the surface

S

is called

smooth.

For a smooth surface

S

, the tangent plane is the plane

containing the vectors

R

u

and

R

v

. Clearly,

R

u

×

R

v

is normal

to the tangent plane.

(63)

If

R

u

×

R

v

6

= 0, the surface

S

is called

smooth.

S

R

u

and

R

v

.

Clearly,

R

u

×

R

v

is normal

(64)

If

R

u

×

R

v

6

= 0, the surface

S

is called

smooth.

S

R

u

and

R

v

. Clearly,

R

u

×

R

v

is normal

(65)

Example

Find the equation of the tangent plane to the surface given by

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

First, note thatP0 is generated whenu=v= 0. Next, we solve the partial derivatives:

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

The normal vector to the tangent plane at (0,−1,1) is

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

Hence, the equation of the tangent plae at (0,−1,1) is

−x+y+ 1 = 0

(66)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

First, note thatP0 is generated whenu=v= 0.

Next, we solve the partial derivatives:

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(67)

Tangent Plane to Parametric Surfaces

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(68)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui

⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(69)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(70)

Tangent Plane to Parametric Surfaces

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i

⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(71)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(72)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i

= det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(73)

Tangent Plane to Parametric Surfaces

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0





=h−1,1,0i

−x+y+ 1 = 0

(74)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,

1,0i

−x+y+ 1 = 0

(75)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,

0i

−x+y+ 1 = 0

(76)

Tangent Plane to Parametric Surfaces

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(77)

Example

x

= sin

v,

y

=

v

−

1

,

z

=

e

u

at the point

P

0

(0

,

−

1

,

1).

Solution.

Ru(u, v) =h0,0, eui ⇒ Ru(0,0) =h0,0,1i

Rv(u, v) =hcosv,1,0i ⇒ Rv(0,0) =h1,1,0i

h0,0,1i × h1,1,0i = det 



ˆı ˆ ˆk 0 0 1 1 1 0



 =h−1,1,0i

−x+y+ 1 = 0

(78)

Exercises

1 _{Identify the surface given by the vector equation}

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

2 _{Find a set of parametric equations for the surface obtained by} revolving the circlex2₊_y2₌_a2 _{about the}_y-axis.

3 _{Find the equation of the tangent plane to the given parametric} surface at the specified point.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

b. R(u, v) =u2_ˆ_ı_{+ 2u}_sin_vˆ_₊_u_cos_v_k;ˆ _u_{= 1, v}_{= 0}

4 Show that the parametric equations

x=asinucosv, y=bsinusinv, z=ccosu,

wherea, bandc are constants, represent an ellipsoid.

5 _{Find a set of parametric equations for the}

torus, obtained by rotating the circle on the xz-plane centered at (b,0,0) about thez-axis. (Hint: Take the parametersθ andα,

as shown.)

(79)

Exercises

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

as shown.)

(80)

Exercises

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

as shown.)

(81)

Exercises

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

as shown.)

(82)

Exercises

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

4 _{Show that the parametric equations}

as shown.)

(83)

Exercises

R(u, v) =hu+v,3−v,1 + 4u+ 5vi.

a. x=u+v, y= 3u2, z=u−v; (2,3,0)

4 _{Show that the parametric equations}

as shown.)

(84)

References

1 _{Stewart, J.,} _{Calculus, Early Transcendentals, 6 ed., Thomson} Brooks/Cole, 2008

2 _{Leithold, L.,}_{The Calculus 7, Harper Collins College Div., 1995}

3 _{Dawkins, P.,}_{Calculus 3, online notes available at} http://tutorial.math.lamar.edu/